1. 1 Permeability

2. 2 Soil Permeability- Definition It is a property of soil that allows the flow of fluid through its interconnected void space OR It is a measure of how easily a fluid (e.g. water) can pass through the soil Different soil has different permeability's.

3. 3 Soil Permeability -Definition Soils consists of solid particles with interconnected voids where water can flow from a point of high energy to a point of low energy water

4. What is Permeability? Permeability is the measure of the soil’s ability to permit water to flow through its pores or voids water Loose soil - easy to flow - high permeability Dense soil - difficult to flow - low permeability

5. 5 Soil Permeability- Definition

6. 6 Importance of Permeability The following applications illustrate the importance of permeability in geotechnical design: The design of earth dams is very much based upon the permeability of the soils used. The stability of slopes and retaining structures can be greatly affected by the permeability of the soils involved. Filters made of soils are designed based upon their permeability.

7. 7 Importance of Permeability Estimating the quantity of underground seepage (Ch. 8) Solving problems involving pumping seepage water from construction excavation

8. 8 Importance of Permeability

9. 9 Why does water flow? If flow is from A to B, the energy is higher at A than at B. water A B Energy is dissipated in overcoming the soil resistance and hence is the head loss.

10. 10 Bernoulli’s Equation 3. Kinetic energy datum z fluid particle The energy of fluid comprise of: 2. Pressure energy 1. Potential energy - due to velocity - due to pressure - due to elevation (z) with respect to a datum

11. Then at any point in the fluid, the total energy is equal to Total Energy = Potential energy + Pressure energy + Kinetic energy =  w gh + P + ½  w v 2 Expressing the total energy as head (units of length) Total Head = Elevation Head + Pressure Head + Velocity Head Bernoulli’s Equation For flow through soils, velocity (and thus velocity head) is very small. Therefore, v 2 /2g = zero. 0

12. Bernoulli’s Equation At any point The head loss between A and B Head loss in non-dimensional form Hydraulic gradient Distance between points A and B Difference in total head

13. PA/wPA/w PB/wPB/w Tricky case!! Remember always to look at total head

14. 14 Hydraulic Gradient In the field, the gradient of the head is the head difference over the distance separating the 2 wells.

15. Datum h A = total head W.T. h = h A - h B W.T. Impervious Soil pervious Soil h B = total head Hydraulic Gradient A B L

16. A B Soil Water In h =h A - h B Head Loss or Head Difference or Energy Loss hAhA hBhB i = Hydraulic Gradient (q) Water out L = Drainage Path Datum hAhA W.T. hBhB ) h = h A - h B W.T. Impervious Soil ZAZA Datum ZBZB Elevation Head Pressure Head Elevation Head Total Head

17. Since velocity in soil is small, flow can be considered laminar v ∝ i v = discharge velocity = i = hydraulic gradient v = k i k = coefficient of permeability Darcy ’ s Law:

18. Since velocity in soil is small, flow can be considered laminar v = k.i Where: v = discharge velocity which is the quantity of water flowing in unit time through a unit gross cross-sectional area of soil at right angles to the direction of flow. k= hydraulic conductivity (has units of L/T) i =hydraulic gradient = h/L Then the quantity of water flowing through the soil per unit time is Discharge = Q = v. A = k (h/L). A Darcy ’ s Law:

19. Datum h A = total head W.T. h = h A - h B W.T. Impervious Soil pervious Soil h B = total head Flow in Soil A B L

20. To determine the quantity of flow, two parameters are needed * k = hydraulic conductivity (how permeable the soil medium) * i = hydraulic gradient (how large is the driving head) k can be determined using 1- Laboratory Testing  [constant head test & falling head test] 2- Field Testing  [pumping from wells] 3- Empirical Equations i can be determined 1- from the head loss and geometry 2- flow net (chapter 8)

21. 21 The hydraulic conductivity k is a measure of how easy the water can flow through the soil. The hydraulic conductivity is expressed in the units of velocity (such as cm/sec and m/sec). Hydraulic Conductivity

22. 22 Hydraulic conductivity of soils depends on several factors: – Fluid viscosity (  ): as the viscosity increases, the hydraulic conductivity decreases – Pore size distribution – Temperature – Grain size distribution – Degree of soil saturation Hydraulic Conductivity It is conventional to express the value of k at a temperature of 20 o C.

23. 23 Hydraulic Conductivity, k Typical Values

24. 24 Laboratory Testing of Hydraulic Conductivity Two standard laboratory tests are used to determine the hydraulic conductivity of soil The constant-head test The falling-head test.

25. 25 Constant Head Test The constant head test is used primarily for coarse-grained soils. This test is based on the assumption of laminar flow (Darcy’s Law apply) From Darcy’s Law Where: Q = volume of water collection A = cross section area of soil specimen T = duration of water collection Then compute:

26. 26 Constant Head Test

27. 27 Constant Head Test

28. 28 Constant Head Test

29. 29 Falling Head Test The falling head test is mainly for fine-grained soils. Simplified Procedure: – Record initial head difference, h 1 at t 1 = 0 – Allow water to flow through the soil specimen – Record the final head difference, h 2 at time t = t 2 a = cross sectional area of standpipe A = cross sectional area of soil L

30. 30 Falling Head Test Calculations: a = cross sectional area of standpipe A = cross sectional area of soil Where: A = cross sectional area of the soil a = cross sectional area of the standpipe h 1 = distance to bottom of the beaker before the test h 2 = distance to bottom of the beaker after the test L = length of the sample t = t 2 -t 1 L Then compute:

31. 31 Falling Head Test Calculations: The above equation is derived assuming: The flow through the standpipe = flow through the soil

32. Falling Head Test Derivation of Falling Head equation

33. Equivalent Hydraulic Conductivity on Stratified Soils Horizontal flow Constant hydraulic gradient conditions Analogous to resistors in series

34. Equivalent Hydraulic Conductivity on Stratified Soils Vertical flow Constant velocity Analogous to resistors in parallel

35. Limitations of Laboratory tests for Hydraulic Conductivity i.It is generally hard to duplicate in-situ soil conditions (such as stratification). ii.The structure of in-situ soils may be disturbed because of sampling and test preparation. iii.Small size of laboratory samples lead to effects of boundary conditions.

36. Permeability Tests using Pumping Wells Used to determine the hydraulic conductivity of soil in the field. During the test, water is pumped out at a constant rate from a test well that has a perforated casing. Several observation wells at various radial distances are made around the test well. Continuous observations of the water level in the test well and in the observation wells are made after the start of pumping, until a steady state is reached. The steady state is established when the water level in the test and observation wells becomes constant.

37. Pumping Well with Observation holes Definitions Aquifer: Soil or rock forming stratum that is saturated and permeable enough to yield significant quantities of water (e.g. sands, gravels, fractured rock)

38. Pumping Well with Observation holes Definitions (cont.) Unconfined Aquifer (water table aquifer) is an aquifer in which the water table forms the upper boundary. Confined Aquifer is an aquifer confined between two impervious layers (e.g. clay).

39. Pumping Well with Observation holes Pumping Well in an Unconfined Aquifer q OR If q, h 1, h 2, r 1, r 2 are known, k can be calculated

40. Pumping Well with Observation holes Pumping Well in a Confined Aquifer q If q, h 1, h 2, r 1, r 2 are known, k can be calculated

41. Darcy & Seepage Velocity Darcy velocity V D is a fictitious velocity since it assumes that flow occurs across the entire cross-section of the sediment sample. Flow actually takes place only through interconnected pore channels (voids), at the seepage velocity V S. A = total area A v voids

42. Darcy & Seepage Velocities From the Continuity Eqn. Q = constant “Pipe running full” means “Inputs = Outputs” Q = A V D = A V V s Where: Q = flow rate A = total cross-sectional area of material A V = area of voids V s = seepage velocity V D = Darcy velocity Since A > A V, and Q = constant, V s > V D Pinch hose, reduce area, water goes faster

43. Darcy & Seepage Velocities From the Continuity Eqn. Q = constant “Pipe running full” means “Inputs = Outputs” Q = A V D = A V V s Where: Q = flow rate A = total cross-sectional area of material A V = area of voids V s = seepage velocity V D = Darcy velocity Since A > A V, and Q = constant, V s > V D Pinch hose, reduce area, water goes faster

44. Darcy & Seepage Velocity: Porosity Q = A V D = A V V s, therefore V S = V D ( A/A V ) Multiplying both sides by the length of the medium (L) divided by itself, L / L = 1 V S = V D ( AL / A V L ) = V D ( Vol T / Vol V ) we get volumes Where: Vol T = total volume Vol V = void volume By definition, Vol v / Vol T = n, the sediment porosity So the actual velocity: V S = V D / n

45. Seepage Velocity (True Velocity)

46. 1) flow lines and equipotentials are at right angles to one another. 2) the cylinder walls are also flow lines. 3) distances between the equipotentials are equal head drops between the equipotentials are also equal. 2.12 Graphical Solutions - Flow Nets Equi-potentials Flow Lines Water IN

47. 2.12 Asymetric Flow Intersections are at right angles approximate to curvilinear square A B C D

48. 2.12 Asymetric Flow Intersections are at right angles approximate to curvilinear square A B C D n d pressure drops a

49. pressure drop between AB and CD is H and let there be n d pressure drops and n f flow lines. 2.12 Asymetric Flow (continued) where q f is the flow per unit cross-section and a x 1 is the cross- section between flow lines. the total seepage =

50. Summary of Flow Nets Solutions are relatively straightforward. 1)draw the appropriate flow net 2)count the number of pressure drops in the flow net (over the relevant distance) 3)count the number of flow lines 4)do a simple calculation work out total flow work out pressure at any given point etc.

51. 51 Flow Net Theory 1. Streamlines  and Equip. lines  are . 2. Streamlines  are parallel to no flow boundaries. 3. Grids are curvilinear squares, where diagonals cross at right angles. 4. Each stream tube carries the same flow.

52. 52 Flow Net in Isotropic Soil Portion of a flow net is shown below   Stream tube

53. 53 Flow Net in Isotropic Soil The equation for flow nets originates from Darcy’s Law. Flow Net solution is equivalent to solving the governing equations of flow for a uniform isotropic aquifer with well-defined boundary conditions.

54. 54 Flow Net in Isotropic Soil Since the flow net is drawn with squares, then d m  dl, and: q = (m/n)KH [L 2 T -1 ] where: q = rate of flow or seepage per unit width m= number of flow channels n= number of equipotential drops h = total head loss in flow system K = hydraulic conductivity

55. 55 Drawing Method: 1. Draw to a convenient scale the cross sections of the structure, water elevations, and aquifer profiles. 2. Establish boundary conditions and draw one or two flow lines  and equipotential lines  near the boundaries.

56. 56 Method: 3. Sketch intermediate flow lines and equipotential lines by smooth curves adhering to right-angle intersections and square grids. Where flow direction is a straight line, flow lines are an equal distance apart and parallel. 4. Continue sketching until a problem develops. Each problem will indicate changes to be made in the entire net. Successive trials will result in a reasonably consistent flow net.

57. 57 Method: 5. In most cases, 5 to 10 flow lines are usually sufficient. Depending on the no. of flow lines selected, the number of equipotential lines will automatically be fixed by geometry and grid layout. 6. Equivalent to solving the governing equations of GW flow in 2-dimensions.

58. 58 Flow Nets: an example A dam is constructed on a permeable stratum underlain by an impermeable rock. A row of sheet pile is installed at the upstream face. If the permeable soil has a hydraulic conductivity of 150 ft/day, determine the rate of flow or seepage under the dam.

59. 59 Flow Nets: an example The flow net is drawn with: m = 5 n = 17

60. 60 Flow Nets: the solution Solve for the flow per unit width: q = (m/n) K h = (5/17)(150)(35) = 1544 ft 3 /day per ft